The permutation (13)(24) acts on vertex 1 by sending it to vertex 3, on vertex 2 sending it to vertex 4, etc.
Example (from lecture)
Let G=D3 and X={1,2,3} representing the three vertices of the triangle
If g=(12) and x=2
Then gx=1 since it sends 2 to 1
If g=(12), h=(13), x=2
(gh)x=(132)2=1
g(hx)=(12)(13)2=(12)2=1
Examples (from discussion)
G×G→G
(g,h)→gh
eh=h
g1(g2h)=(g1g2)h
GLn(R)×Rn→Rn
(A,v)→Av
Inv=v
A⋅(Bv)=(AB)v
In general, if X is any set and G is a subgroup of SX (i.e. the group of all permutations acting on X), then X is a G-set under the group action (σ,x)↦σ(x) for σ∈G and x∈X
Difference between left and right actions
For right action, we would write action as xg, with the variable on the left and function on the right
In typical math notation, the function that acts on the variable should be the function that’s closest to the variable
So for xgh, g would first act on x, then h acts on xg
Left and right actions are not interchangable because
Let ghx=xhg
Then ghx=(gh)x=x(hg)=xhg
This implies gh=hg which is not true unless G is abelian
Three objects regarding group actions
Fixed point set
Let group G act on set X, the fixed point set of g in X, denoted Xg, is Xg={x∈X∣gx=x}, where g is fixed
Xg⊆X
Example
For G=D3, X={1,2,3}
X(12)=3
X(123)=∅
Xe={1,2,3}
Stabilizer subgroup/isotropy subgroup
Let group G act on set X, the stabilizer subgroup x, denoted Gx, is Gx={g∈G∣gx=x}, where x is fixed
Gx is a subgroup of G
Example
G={e,(12),(34),(12)(34)}≤S5
X={1,2,3,4,5}
G1={e,(34)}
G5=G
Orbit
Let group G act on set X, the orbit of an element x∈X, denoted Ox is the set of results to gx for all g∈G
The type of outcome of στσ−1 always is the same type as τ when σ,τ∈Sn
In the above examples, when τ is a 3-cycle, στσ−1 is a 3-cycle; when τ is a 2-2-cycle, στσ−1 is a 2-2-cycle
How to calculate στσ−1 the fast way when σ,τ∈Sn
We know the result is the same type as τ
Take every individual number from τ, see where that number maps to in σ, that resulting mapped number is in the same location as where the number originated from in τ
Normal subgroups and conjugation in Sn
K4={e,(12)(34),(13)(24),(14)(23)}⊆S4
In order to check H is a normal subgroup in G, we need to check ghg−1∈H for all h∈H, g∈G
K4 is a normal subgroup in S4
{e,(12)} isn’t normal because ∃σ,σ(12)σ−1=(32)
From the fast way above, we know σ=(13)
Therefore we know a subgroup of Sn is normal if all elements of the subgroup is the identity + all cycles that are the same type
Conjugation in Q8
Let G=Q8={±1,±i,±j,±k}
We define i,j,k to be:
i2=j2=k2=−1
ij=k,jk=i,ki=j
ji=−k,kj=−i,ik=j
Example conjugation
kik−1
To figure out what k−1 is
k4=1=k0⟹k−1=k3=k2k=−k
So kik−1=(ki)k−1=jk−1=jk3=j(−k)=−jk=−i
Conjugation class and orbits
Let’s fix x∈G
Ox={gx∣g∈G}={gxg−1∣g∈G} when in terms of conjugation
This above would be called a conjugacy class
Conjugacy class - all elements ∈G that are conjugate to x∈G (i.e. orbit of x∈G)
Example - conjugacy class of Q8
O1={1}
O1={g1g−1}={gg−1}={1}
O−1={−1}
Oi={i,−i}
Oj={j,−j}
Ok={k,−k}
Conjugation, fixed point set, stabilizer group example (from discussion)
We define ZG(X)={g∈G∣gx=xg,∀x∈X}
H×G→G
(h,g)→hgh−1
Find the fixed point set Xh and stabilizer group Gg
Xh={g∈G∣hgh−1=g}={g∈G∣hg=gh}=ZG(h)
Gg={h∈H∣hgh−1=g}={h∈H∣hg=gh}
Gg=ZH(g) because g might not be in H, so ZH(g) doesn’t make sense as a group
Apr 5 Textbook Notes
Conjugacy classes
Let X be a finite G-set and XG be the set of fixed points in X
XG={x∈X:gx=x for all g∈G}
Since the orbits of the action partition X, we have ∣X∣=∣XG∣+∑i=kn∣Oxi∣, where xk,⋯,xn are representatives from distinct nontrivial orbits of X
Consider the special case in which G acts on itself by conjugation, (g,x)↦gxg−1
The center of G, Z(G)={x:xg=gx for all g∈G}, is the set of points that are fixed by conjugation
The nontrivial orbits of the action are called the conjugacy classes of G
If x1,⋯,xk are representatives from each of the nontrivial conjugacy classes of G and ∣Ox1∣=n1,⋯,∣Oxk∣=nk, then
∣G∣=∣Z(G)∣+n1+⋯+nk
The stabilizer subgroups of each of the xi’s, C(xi)={g∈G:gxi=xig}, are called the centralizer subgroups of xi’s
From theorem 14.11, we obtain the class equation
∣G∣=∣Z(G)∣+[G:C(x1)]+⋯+[G:C(xk)]
Essentially what this means is that ∣X∣ is equal to ∑∣Ox∣, the sum of cardinality of distinct orbits
One of the consequences of the class equation is that the order of each conjugacy class must divide the order of G
Examples (from textbook)
For S3
The conjugacy classes in S3 are the following
{(1)}
{(123},(132)}
{(12),(13),(23)}
The class equation is 6=1+2+3
For D4
The center of D4 is {(1),(13)(24)}
The conjugacy classes are
{(13),(24)}
{(1432)(1234)}
{(12)(34),(14)(23)}
The class equation for D4 is 8=2+2+2+2
Theorem 14.15
Let G be a group of order pn, where p is prime, then G has a nontrivial center
Chapter 16 - Rings
16.1 - Rings
Apr 7 Lecture + Textbook Notes
Ring
A nonempty set R is a ring if it has two closed binary operations, addition and multiplication, satisfying the following conditions:
For a,b∈R, a+b=b+a
For a,b,c∈R, (a+b)+c=a+(b+c)
There’s an element 0 in R such that a+0=a for all a∈R
For every a∈R, there exists an element −a∈R such that a+(−a)=0
For all a,b,c∈R, (ab)c=a(bc)
For all a,b,c∈R
a(b+c)=ab+ac
(a+b)c=ac+bc
(Multiplicative) identity/unity - an element, denoted 1∈R, such that 1=0 and 1a=a1=a for each element a∈R
Rings are NOT required to have a multiplicative identity
Multiplicative inverse - an element in a−1∈R such that aa−1=a−1a=1
Rings are NOT required to have a multiplicative inverse
Unit - a nonzero element of a ring that has a multiplicative inverse
Commutative ring - a ring R for which ab=ba for all a,b∈R
Integral domain - a commutative ring with a multiplicative identity where if ab=0 for every a,b∈R, either a=0 or b=0
i.e. it’s a commutative ring with a 1 with no zero divisors
Division ring - a ring R, with an identity, in which every nonzero element in R is a unit
i.e. division ring is a ring with an identity where, for every a∈R,a=0, there exists a unique element a−1 where a−1a=aa−1=1
Field - a commutative division ring
Zero divisor (of commutative ring R) - a nonzero element a in a commutative ring R when there’s exists a nonzero element b∈R such that ab=0
Examples of rings
(Z,+,⋅)
If ab=0 then a=0 or b=0, so it’s a ring
It’s not a field because there’s no multiplicative inverse of elements others than −1 and 1
(Q,+,⋅)
(R,+,⋅)
(C,+,⋅)
The three above are all both rings and fields
All numbers except 0 is a unit
(Zn,+,⋅)
It’s a commutative ring
Z4 is not an integral domain because (3⋅4)mod12≡0 in Z12
So 3 and 4 are zero divisors in Z12
Z3 is an intergral domain
Z2 is a field
M2(R)
2×2 matrices with real entries under usual addition and multiplication form a noncommmutative ring, since AB=BA and we can have AB=0 when neither A or B is zero
A matrix is a unit if it has a non-zero determinant (i.e. invertible)
H={a+bi+cj+dk∣a,b,c,d∈R,i,j,k∈Q8}
i.e. quaternion algebra
It’s a division ring
Proposition 16.8
Let R be a ring with a,b∈R, then
a0=0a=0
a(−b)=−a(b)=−ab
(−a)(−b)=ab
Proof
0+0=0⟹a(0+0)=a0⟹a0+a0=a0⟹a0=0
ab+a(−b)=a(b−b)=a0=0
So −ab=a(−b)
Similarly, −ab=(−a)b
(−a)(−b)=−(a(−b))=−(−ab)=ab
Proposition
If A∈M2(Z) is a unit, then det(A)=±1
Proof
Let A be a unit
So det(A) must be a factor of a,b,c,d
Also det(A)=0
Let det(A)=ad−bc=k=0
Then k∣a and k∣d⟹k2∣ad
For the same reasoning, k2∣bc
⟹k2∣(ad−bc)
⟹k2∣k
Only possible when k=±1
k=0
Subring
Subring - a subring S of ring R is a subset S of R such that S is also a ring under the inherited operations from R
Example of subring
The ring nZ is a subring of Z
Chain of subrings
Z⊂Q⊂R⊂C
Proposition 16.10
Let R be a ring and S a subset of R, S is a subring of R if and only if the following conditions are satisfied
S=∅
For all r,s∈S, rs∈S
For all r,s∈S, r−s∈S
Example
Let R=M2(R) be the ring of 2×2 matrices with real entries
And let T be the set of upper triangular matrices in R
T={(a0bc)}
T is a subring of R
Let A=(a0bc) and B=(a′0b′c′)
AB=(aa′0ab′+bc′cc′)∈T
A−B=(a−a′0b−b′c−c′)∈T
16.2 - Integral domains and fields
Apr 10 Lecture + Textbook Notes
Definitions
Zero divisor - a nonzero element r in a commutative ring R such that there’s some nonzero element s such that rs=0
Integral domain - a commutative ring with an identity with no zero divisors
Unit - an element a, in a ring R with identity, with a multiplicative inverse
Division ring - a ring where every nonzero element is a unit
Field - a commutative division ring
Example of integral domain (or not)
Let R2=R×R such that (a,b)⋅(c,d)=(ac,bd) for (a,b),(c,d)∈R2
It’s commutative
It has an identity (1,1)
It has zero divisors (e.g. (0,1)⋅(1,0))
So it’s not an integral domain
Examples of integral domain
Z
All fields
Z[i]={a+bi∣a,b∈Z}
R[x]={polynomial functions ∣a,b∈R}
Zp, where p is prime
Z3[i]
Examples of not integral domain
R2=R×R
Z4
Z6
H
Zn, where n is a composite
Example of units
Let i2=−1, then the set Z[i]={a+bi:a,b∈Z} forms a ring called the Gaussian integers
The Gaussian integers are a subring of complex numbers closed under addition and multiplication
Let α=a+bi be a unit in Z[i], then α=a−bi is also a unit since, if αβ=1, then αβ=1
If β=c+di, then
1=αβαβ=(a2+b2)(c2+d2)
Therefore a2+b2 must be either −1 or 1 (i.e. a+bi=±1 or ±i)
Therefore units of this ring are ±1 and ±i
Hence the Gaussian integers are not a field
Cancellation law
Proposition 16.15 - cancellation law
Let D be a commutative ring with identity, then D is an integral domain ⟺for all nonzero elements a∈D with ab=ac, we have b=c
Proof
Let D is an integral domain, then D has no zero divisors
Let ab=ac with a=0, then ab−ac=0
a(b−c)=0
Since a=0, and there are no zero divisors, then b−c=0
b=c
Conversely, let’s suppose cancellation is possible in D (i.e. ab=ac⟹b=c)
Let ab=0 and a=0
Then ab=a0⟹b=0
Therefore a can’t be a zero divisor
Corollary
In an integral domain, if xy=0, then x=0 or y=0
Integral domains and fields
Proposition
If x∈R is a unit, then x is not a zero divisor
Proof
Let x∈R be a unit, and there exists y∈R such that xy=1=yx
We know that x=0
If x is a zero divisor, then there exists z∈R, z=0, such that xz=0
Then y(xz)=yxz=0
1⋅z=0
⟹z=0
This contradicts that z=0
So x cannot be a zero divisor
Corollary
All fields are integral domains
Characteristic of a ring
For any nonnegative integer n and any element r in ring R, we write r+⋯+r (n times) as nr
Characteristic (of a ring) (denoted charR) - the least positive integer n such that nr=0 for all r∈R; if no such integer exists, then the characteristic of R is defined to be 0
Example
For every prime p, Zp is a field of characteristic p
We know every element in Zp has an inverse so Zp is a field
If a is any nonzero element in the field, then pa=0, since the order of any nonzero element in the abelian group Zp is p
Example (for integral domain)
If R is an integral domain
char(R)={p, where p is prime0
Proof
If char(R)=n=0, n⋅1R=0
Let n=n1⋅n2, such that n1,n2∈R and n1,n2=0
(n1⋅1R)(n2⋅1R)=n1n2⋅(1R1R)=n1R=0
Then n1 and n2 are zero divisors, so R isn’t an integral domain
Lemma 16.18
Let R be a ring with identity
If 1 has order n, then the characteristic of R is n
Theorems
Theorem 16.16
Every finite integral domain is a field
Proof
Let D be a finite integral domain and D∗ be the set of nonzero elements of D
We must show that every element in D∗ has an inverse
For each a∈D∗ we can define a map λa:D∗→D∗ by λa(d)=ad
This map makes sense because if a=0 and d=0, then ad=0
This map is one-to-one, since for d1,d2∈D∗,
ad1=λa(d1)=λa(d2)=ad2
which implies d1=d2 by left cancellation
Since D∗ is a finite set, the map λa must also be onto
Hence, for some d∈D∗, λa(d)=ad=1
Therefore, a has a left inverse
Since D is commutative, d be also have a right inverse for a
Consequently, D is a field
Theorem 16.19
The characteristic of an integral domain is either prime or zero
Proof
Let D be an integral domain and suppose that the characteristic of D is n with n=0
If n isn’t prime, then n=ab, where 1<a,b<n
By lemme 16.18, we need to only consider the case n1=0
Since 0=n1=(ab)1=(a1)(b1) and there are no zero divisors in D, either a1=0 or b1=0
Hence, the characteristic of D must be less than n, which is a contradiction
Therefore n must be prime
Unit group (from discussion)
Define R∗={a∈R∣a is a unit}
Examples
Z∗={±1}
Q=Q∖{0}
Given a field F
F∗=F∖{0}
16.3 - Ring homomorphisms and ideals
Apr 12 Lecture + Textbook Notes
Ring homomorphisms
If R and S are rings, then a ring homomorphism is a map ϕ:R→S satisfying the following for all a,b∈R
ϕ(a+b)=ϕ(a)+ϕ(b)
ϕ(ab)=ϕ(a)ϕ(b)
If ϕ:R→S is a one-to-one and onto homomorphism, then ϕ is a isomorphism of rings
For any ring homomorphism ϕ:R→S, the kernel of the ring homomorphism is the set
ker(ϕ)={r∈R:ϕ(r)=0}
Example
ϕ:Z→Zn by a↦amodn
ϕ(a+b)=(a+b)modn=amodn+bmodn=ϕ(a)+ϕ(b)
ϕ(ab)=(ab)modn=(amodn)(bmodn)=ϕ(a)ϕ(b)
kerϕ=nZ
Example that’s not homomorphism
Z[5]→Z[7] is not a ring homomorphism
Doesn’t preserve multiplication
ϕ(5⋅5)=ϕ(5)=5
ϕ(5)ϕ(5)=77=7=ϕ(5⋅5)
Proposition 16.22
Let ϕ:R→S be a ring homomorphism
If R is a commutative ring, then ϕ(R) is a commutative ring
ϕ(0)=0
Let 1R and 1S be the identities for R and S respectively. If ϕ is onto, then ϕ(1R)=1S
If R is a field and ϕ(R)={0}, then ϕ(R) is a field
Ideals
Ideal - a subring I of ring R such that if i∈I and r∈R, then ir∈I and ri∈I (i.e. it could also be a subgroup under addition, together with multiplication by R (kind of like “R-action”))
rI is a left ideal
Ir is a right ideal
Two-sided ideal - when an ideal I of ring R has rI⊂I and Ir⊂I for all r∈R
In a commutative ring, any ideal is two-sided
One-sided ideal - when an ideal I of ring R has rI⊂I or Ir⊂I for r∈R, but not both
(x)={rx∣r∈R}
(x,y)={r1x+r2y∣r1,r2∈R}
(x1,⋯,xn)={∑i=1nrix1∣ri∈R}
Examples of ideals
{0} and R are trivial ideals
R=Z
(n)= ideal generated by n, n≥0
(4)⊊(2), these both are ideals of Z
(0)={0}, the only finite ideal inside Z
R=R[x]
I=(x+1)
Let r=x∈R
x(x+1)=x2+x∈I
Let r=x−1
(x−1)(x+1)=x2−1∈I
So (x+1)={f(x)(x+1)∣f(x)∈R[x]}
I={g(x)∣g(−1)=0}
R=Z6
(2)={0,2,4}
R=R[x,y]
I=(x,y)={f(x,y)x+g(x,y)y∣f,g∈R[x,y]}
x+y∈I
(x+1)x∈I
(x+1)y∈I
1∈/I, because no constant term can be in I
xy−2∈/I, because of the −2 being constant
0∈I
From this we can conclude that I isn’t a principal ideal
Proof
Assume, for contradiction, I is principal such that I=(f)⟹f∣x and f∣y
Such a polynomial f doesn’t exist
To prove a subring I of ring R is an ideal
Prove that I is a group under addition
Show that ri∈I for all r∈R, i∈I
Principal ideal - the ideal generated by element a∈R, denoted (a), where R is a commutative ring with identity
(a) is nonempty since both 0=a0 and a=a1 are in (a)
ar+ar′=a(r+r′)
−ar=a(−r)
r′(ar)=a(r′r)
Theorem 16.25
Every ideal in a ring of integers Z is a principal ideal
Proof
The zero ideal is a principal ideal since (0)={0}
If I is any nonzero ideal in Z, then I must contain some positive integer m
So there exists a least positive integer n in I by the principle of well-ordering
Now let a be any element in I
Using the division algorithm, we know that there exists integers q and r such that a=nq+r, where 0≤r<n
This equation tells us that r=a−nq∈I, but r must be 0 since n is the least positive element in I
Therefore a=nq and I=(n)
Proposition 16.27
The kernel of any ring homomorphism ϕ:R→S is an ideal in R
Proof
We know from group theory that kerϕ is an additive subgroup of R
Suppose that r∈R and a∈kerϕ, need to show that ar and ra are in kerϕ
ϕ(ar)=ϕ(a)ϕ(r)=0ϕ(r)=0
ϕ(ra)=ϕ(r)ϕ(a)=ϕ(r)0=0
Claim: 1R∈I⊆R⟹I=R (from discussion)
Proof
1R∈I⟹r⋅1R∈I,∀r∈R⟹I=R
Apr 14, 17 Lecture + Textbook Notes
Quotient rings
Theorem 16.29
Let I be a two-sided ideal of R, the factor group R/I is a ring with multiplication defined by (r+I)(s+I)=rs+I
Proof
We already know that R/I is an abelian group under addition
Let r+I,s+I∈R/I, we must show that (r+I)(s+I)=rs+I is independent of choice of coset
i.e. if r′∈r+I and s′∈s+I, then r′s′∈rs+I must be true
Since r′∈r+I and s′∈s+I, there exists an element a,b∈I such that r′=r+a and s′=s+b
r′s′=(r+a)(s+b)=rs+as+rb+ab
We know as+rb+ab∈I, since I is an ideal
Therefore r′s′∈rs+I
The ring R/I is the factor or the quotient ring
Example
Let R=Z, I=(3)
Z/(3)=Z/3
Let R=R[x], I=(x2+1)
f(x)+(x2+1)∈R[x]/(x2+1)
f(x) can always be written as degree 0 or 1
e.g. x3+(x2+1)=−x+(x2+1)
Proof
If we take the difference of x3 and −x, we get
x3+x+(x2+1)=0+(x2+1)
We know x2+1 divides x3+x
x2+1∣x3+x
Therefore x3+(x2+1)=−x+(x2+1)
−x5+(x2+1)=−x+(x2+1)
Proof
x5−x=(x2+1)(x3−x)⟹x2+1∣x5−x
So −x5+(x2+1)=−x+(x2+1)
OR −x5+(x2+1)=x3+(x2+1)
So −x5+(x2+1)=−x+(x2+1)
(x+I)2=x2+(x2+1)=−1+(x2+1)
Theorem 16.30
Let I be an ideal of R, the map ϕ:R→R/I (called the natural homomorphism or canonical homomorphism), defined by ϕ(r)=r+I is a ring homomorphism of R onto R/I with kernel I
Proof
ϕ:R→R/I is certainly a surjective abelian group homomorphism
We only need to show that ϕ works correctly under ring multiplication
Let r,s∈R
ϕ(r)ϕ(s)=(r+I)(s+I)=rs+I=ϕ(rs)
Isomorphism and correspondence theorems
Theorem 16.31 (First Isomorphism Theorem)
Let ψ:R→S be a ring homomorphism
Then kerψ is an ideal of R
If ϕ:R→R/kerψ is the canonical homomorphism, then there exists a unique isomorphism η:R/kerψ→ψ(R) such that ψ=ηϕ
Proof
Let K=kerψ
By the First Isomorphism Theorem for groups, there exists a well-defined group homomorphism η:R/K→ψ(R) defined by η(r+K)=ψ(r) for additive abelian groups R and R/K
To show this is a ring homomorphism, we only need to show that η((r+K)(s+K))=η(r+K)η(s+K)
Let I be a subring of a ring R and J be an ideal of R, then I∩J is an ideal of I and I/(I∩J)≅(I+J)/J
Theorem 16.33 (Third Isomorphism Theorem)
Let R be a ring and I and J be ideals of R where J⊂I, then R/I≅I/JR/J
Theorem 16.34 (Correspondence Theorem)
Let I be an ideal of ring R, then S↦S/I is a one-to-one correspondence between the set of subrings S containing I and the set of subrings of R/I
Furthermore, the ideals of R containing I corresponds to the ideals of R/I
i.e. if there’s an ideal in R/I, there’s an ideal J where I⊆J⊆R, such that J↦J/I
Examples (from lecture)
R=Z, R/I=Z6
I=(6)
Ideals in Z6: {0},{0,3},{0,2,4},Z6
J is ideals in Z containing (6)
J can be Z,(2),(3),(6)
When J=(2), in Z6, J corresponds to {0,2,4}
When J=(3), J corresponds to {0,3}
J=Z, Z6
J=(6), {0}
R=Z,R/I=Z12
I=(12)
Ideals in Z12=Z12,{0,2,4,6,8,10},{0,3,6,9},{0,4,8},{0,6},{0}
J can be Z,(2),(3),(4),(6),(12)
J=Z⟺Z12
J=(2)⟺{0,2,4,6,8,10}
J=(3)⟺{0,3,6,9}
J=(4)⟺{0,4,8}
J=(6)⟺{0,6}
J=(12)⟺{0}
R=R[x], I=(x2−1)
f↔(f(1),f(−1))
R/I=R[x]/(x2−1)
J need to be a ideal generated by a factor of x2−1 to contain (x2−1)
Therefore J can be (1)=R[x],(x−1),(x+1),(x2−1)
Ideals in R[x]/(x2−1):
Trivially, {0} and R[x]/(x2−1)
(x−1)/(x2−1)
(x+1)/(x2−1)
R[x]/(x2−1)↔R2
(x−1)/(x2−1)↔(0, any)
(x+1)/(x2−1)↔(any,0)
(0)↔(0,0)
16.4 - Maximal and prime ideals
Apr 19 Lecture + Textbook Notes
Maximal and prime ideals
Maximal ideal (of ring R) - a proper ideal M of ring R such that M isn’t a proper subset of any ideal of R except R itself
Theorem 16.35
Let R be a commutative ring with identity and M be an ideal in R, then M is a maximal ideal of R⟺R/M is a field
Proof
⟹
Let M be a maximal ideal in R
If R is a commutative ring, then R/M must also be a commutative ring
1+M acts as an identity for R/M
So we just need to show that every nonzero element in R/M has an inverse
If a+M is a nonzero element in R/M, then a∈/M
Define I={ra+m:r∈R and m∈M}
We’ll show I is an ideal in R
The set I is nonempty since 0a+0=0 is in I
If r1a+m1 and r2a+m2 are two elements in I, then (r1a+m1)−(r2a+m2)=(r1−r2)a+(m1−m2) is in I
Also, for any r∈R, it’s true that rI⊂I
Hence, I is closed under multiplication and satisfies the necessary conditions to be an ideal
Therefore I is an ideal properly containing M
Since M is a maximal ideal, I=R
Consequently, by the definition of I, there must be an m∈M and b∈R such that 1=ab+m
Therefore 1+M=ab+M=ba+M=(a+M)(b+M)
⟸
Suppose M is an ideal and R/M is a field
Since R/M is a field, it must contain at least two elements: 0+M=M and 1+M
Hence M is a proper ideal of R
Let I be any ideal properly containing M
We need to show that I=R
Choose a∈I but a∈/M
Since a+M is a nonzero element in a field, there exists an element b+M∈R/M such that (a+M)(b+M)=ab+M=1+M
Consequently, there exists an element m∈M such that ab+m=1, and 1∈I
Therefore, r1=r∈I for all r∈R
Consequently I=R
Prime ideal - a proper ideal P in a commutative ring R where when ab∈P, then either a∈P or b∈P
Examples
Let R=Z, I=pZ, where p is prime
pZ is a maximal ideal since Z/pZ≅Zp is a field
If xy∈pZ, then p∣x or p∣y
So pZ=(p) is also a prime ideal
Let R=Z12, P={0,2,4,6,8,10}
P is a prime ideal and a maximal ideal
Let R=Z, I=(0)
I is a prime ideal
If xy=0, then x=0 or y=0
Let R=R2, I=(0)
I is NOT a prime ideal
Because R2 has zero divisors
(0,1)(1,0)=0
Let R=Z4
(2)={0,2} is a maximal and prime ideal
Let R=Z6
(2)={0,2,4} is a maximal and prime ideal
(3)={0,3} is a maximal and prime ideal
Proposition
(0)⊆R is prime ⟺R has no zero divisors when R is a commutative ring with an identity ⟺R is an integral domain
Proposition 16.38
Let R be a commutative ring with identity 1, where 1=0, then P is a prime ideal in R⟺R/P is an integral domain
Proof
⟹
Assume P is an prime ideal in R and (a+P),(b+P)∈R/P such that (a+P)(b+P)=0+P=P
Then ab∈P
If a∈/P, then b must be in P by the definition of a prime ideal
Hence, b+P=0+P and R/P is an integral domain
⟸
Assume P is an ideal in R and R/P is an integral domain
Suppose ab∈P
If a+P and b+P are two elements of R/P such that (a+P)(b+P)=0+P=P, then either a+P=P or b+P=P
This means either a∈P or b∈P
Therefore P must be prime
Example
Every ideal in Z is in the form nZ
The factor ring Z/nZ≅Zn is an integral domain only when n is prime
It’s actually a field
Hence, the nonzero prime ideals in Z are the ideals pZ, where p is prime
Corollary 16.40
Every maximal ideal in a commutative ring with identity is also a prime ideal
Proof (incomplete)
Let M⊂R be a maximal ideal
The correspondence theorem says that the ideals in R/M corresponds to the ideals J in R such that M⊆J⊆R
Since M is a maximal ideal, the only ideals that satisfy M⊆J⊆R are M and R
That corresponds to the ideals (0) and R/M in R/M
Proposition
For maximal ideal M of commutative ring with identity R, R/M is a field
Proof
Let x∈R/M, x=0
The ideal (x) is not (0)
Thus (x)=R/M
Now 1∈R/M=(x)
So there exists y∈R/M such that xy=1
Therefore x is a unit
16.5 - An application to software design
Apr 21 Lecture + Textbook Notes
Chinese Remainder Theorem
Lemma 16.41
Let m and n be positive integers such that gcd(m,n)=1, then for a,b∈Z, the following system has a solution
x≡amodm
x≡bmodn
If x1 and x2 are two solutions of the system, then x1≡x2(modmn)
Proof
Example
Solve the system
x≡3mod4
x≡4mod5
Using the Euclidean algorithm, we can find integers s and t such that 4s+5t=1
Two such integers are s=4 and t=−3
Consequently
x=a+k1m=3+4k1=3+4[(5−4)4]=19
Theorem 16.43 (Chinese Remainder Theorem)
Let n1,n2,⋯,nk be positive integers (that are ≥2) such that gcd(ni,nj)=1 for i=j, then for any integers a1,⋯,ak, the following system has a solution
x≡a1(modn1)
x≡a2(modn2)
⋯
x≡ak(modnk)
where it’s obtained by x≡A(modN)
where N=n1n2⋯nk, and A is unique
Furthermore, any two solutions of the system are congruent modulo n1n2⋯nk
Proof
Examples (from textbook)
Examples (from lecture)
⎩⎨⎧x≡2(mod3)x≡3(mod5)x≡4(mod7)
Let’s first look at x≡4(mod7)
x∈{4,11,18,25,32,⋯,7n+4∣n∈Z+}
The first number in the list that fulfills x≡3(mod5) is x=18
Let’s now look at x≡18mod5⋅7=18mod35
x∈{18,53,88,⋯,35n+18∣n∈Z+}
First number that fulfills x≡2mod3 is x=53
x≡53mod105 is the solution
⎩⎨⎧x≡3(mod4)x≡4(mod5)x≡2(mod7)
x≡2mod7
x∈{2,9,16,⋯}
First number that fulfills x≡4mod5 is x=9
x≡9mod35
x∈{9,44,79,⋯}
First number that fulfills x≡3mod4 is x=79
x≡79mod140 is the solution
Chapter 17 - Polynomials
17.1 - Polynomial rings
Apr 21 Textbook Notes
Definitions for polynomials
Polynomial over R with indeterminate x - any expression of the form f(x)=∑i=0naixi=a0+a1x1+a2x2+⋯+anxn, where ai∈R and an=0
Coefficients (of f) - a0,a1,⋯,an in f
Leading coefficient - an in f
Monic - when a polynomial’s leading coefficient is 1
Degree (of f) - where n is the largest nonnegative number such that an=0, denoted degf(x)=n
If f=0 is the zero polynomial, then degf(x)=−∞
The set of all polynomials with coefficients in a ring R is denoted R[x]
Let R be a commutative ring with identity, then R[x] is a commutative ring with identity
Proof
Proposition 17.4
Let p(x) and q(x) be polynomials in R[x], where R is an integral domain
Then degp(x)+degq(x)=deg(p(x)q(x))
Furthermore, R[x] is an integral domain ⟺R is an integral domain
Proof
Example from lecture
Let R=Z6
(2x+1)(3x2+2)=6x3+3x2+4x+2=3x2+4x+2
The ring of polynomials in two indeterminates x and y with coefficients in R (denoted R[x,y]) - a ring of polynomials with two variables x,y
The ring of polynomials in n indeterminates with coefficients in R (denoted R[x1,⋯,xn]) - a ring of polynomials with n variables x1,⋯,xn
Theorem 17.5
Let R be a commutative ring with identity and α∈R, then we have a ring homomorphism ϕα:R[x]→R (called the evaluation homomorphism) defined by
ϕα(p(x))=p(α)=anαn+⋯+a1α+a0
where p(x)=anxn+⋯+a1x+a0
Proof
Proposition
deg(f+g)≤max(degf,degg)
17.2 - Division algorithm
Apr 28 Lecture + Textbook Notes
Division algorithm
Theorem 17.6 (Division algorithm)
Let f(x) and g(x) be polynomials in F[x], where F is a field and g(x) is a nonzero polynomial, then there exists unique polynomials q(x),r(x)∈F[x] such that
f(x)=g(x)q(x)+r(x)
where either degr(x)<degg(x) or r(x) is the zero polynomial
Examples
Example (from lecture)
Let R=F5[x], where F5 denotes Z5 because Zp for prime p are fields
f(x)=x3−x−1, g(x)=2x2−1
Then q(x)=3x and r(x)=2x−1
Let p(x) be a polynomial in F[x] and α∈F, α is a zero or root of p(x) is p(x) is in the kernel of the evaluation homomorphism ϕα
Aka p(α)=0⟹α is a zero of p(x)
Corollary 17.8
Let F be a field, an element α∈F is a zero of p(x)∈F[x]⟺x−α is a factor of p(x) in F[x]
Proof
⟹
Suppose α∈F and p(α)=0
By the division algorithm, there exist polynomials q(x) and r(x) such that p(x)=(x−α)q(x)+r(x) and degr(x)<deg(x−α) must be true
Since the degree of r(x) is less than 1, and r(x)=a for a∈F, we have that
p(x)=(x−α)q(x)+a
But 0=p(α)=0⋅q(α)+a=a, consequently, p(x)=(x−α)q(x), and x−α is a factor of p(x)
⟸
Suppose that x−α is a factor of p(x), where p(x)=(x−α)q(x), then p(α)=0⋅q(α)=0
Corollary 17.9
Let F be a field, a nonzero polynomial p(x) of degree n in F[x] can have at most n distinct zeros in F
Proof
Let F be a field
A monic polynomial (i.e. polynomial where the term with the degree n has coefficient 1) d(x) is a greatest common divisor of polynomials p(x),q(x)∈F[x] if d(x) evenly divides both p(x) and q(x)
If for any other polynomial d′(x) dividing both p(x) and q(x), d′(x)∣d(x)
We write d(x)=gcd(p(x),q(x))
Two polynomials p(x),q(x) are relatively prime if gcd(p(x),q(x))=1
Proposition 17.10
Let F be a field and suppose that d(x) is a greatest common divisor of two polynomials p(x),q(x)∈F[x], then there exist polynomials r(x) and s(x) such that
d(x)=r(x)p(x)+s(x)p(x)
Furthermore, the greatest common divisor of two polynomials is unique
Proof
Euclidean algorithm
Example
R=F3[x]
f(x)=x4−x2+1, g(x)=x3+x2+x+1
Find gcd of f and g
f(x)=(x−1)g(x)+(2x2+2)
q(x)=x−1
r(x)=2x2+2
g(x)=(2x+2)r(x)+(0)
q2(x)=2x+2
r2(x)=0
⟹gcd(x4−x2+1,x3+x2+x+1)=2x2+2
f(x)=(2x2+2)(2x2+2)=4(x2+1)2=(x2+1)2
g(x)=(2x2+2)(2x+2)=(x2+1)(x+1)
x2+1 is the monic polynomial that’s the gcd(x4−x2+1,x3+x2+x+1)
17.3 - Irreducible polynomials
May 1 Lecture + Textbook Notes
Irreducible polynomials
Irreducible polynomial - a non-constant polynomial f(x)∈F[x] over a field F where f(x) can’t be expressed as a product of two polynomials g(x),h(x)∈F[x], where degg(x)<degf(x) and degh(x)<degf(x)
They’re like the “prime number” of polynomial rings
Examples of irreducible polynomials (textbook)
x2−2∈Q[x]
It can’t be factors any further over the rational number
x2+1∈Q[x]
p(x)=x3+x2+2∈Z3[x]
Suppose p(x) is reducible over Z3[x]
By the division algorithm, there would have to be a factor of the form x−a, where a is some element in Z3[x]
Hence ∃a∈Z3,p(a)=0
But
p(0)=2
p(1)=1
p(2)=2
Therefore p(x) has no zeros in Z3 and must be irreducible
Examples of reducible/irreducible polynomials (lecture)
x2−1∈Q[x] is reducible
x2−1=(x+1)(x−1)
x2+1∈R[x] is irreducible
x2+1∈C[x] is reducible
x2+1=(x+i)(x−i)
x4+2x2+1∈Q[x] is reducible
x4+2x2+1=(x2+1)2
f(x)=x3−x−1∈F3[x]
f(0)=−1
f(1)=−1
f(2)=−1
So f(x) has no root ⟹f(x) is irreducible
Note
A polynomial being reducible in F[x] doesn’t imply it has roots in F[x]
Example
x4+2x2+1∈Q[x] is reducible
x4+2x2+1=(x2+1)2
But it has no roots in Q
How to tell if a polynomial is irreducible or not
degf(x)=1⟹f(x) is always irreducible
f(x) has a root and degf(x)≥2⟹f(x) is always reducible
If degf(x)≤3, f(x) is reducible ⟺f(x) has a root, f(x) is irreducible ⟺f(x) has no roots
Gauss’s lemma
Eisenstein’s criterion
Rational root theorem
Lemma 17.13
Let p(x)∈Q[x], then p(x)=sr(a0+a1x+⋯+anxn), where r,s,a0,⋯,an are integers, the ai’s are relatively prime, and r and s are relatively prime
Proof
Theorem 17.14 (Gauss’s lemma)
Let p(x)∈Z[x] be a monic polynomial such that p(x) factors into a product of two polynomials — α(x),β(x)∈Q[x] — where the degrees of α(x),β(x) are less than the degree of p(x)
Then p(x)=a(x)b(x)
where a(x),b(x) are monic polynomials in Z[x] with degα(x)=dega(x) and degβ(x)=degb(x)
Proof
Corollary 17.15
Let p(x)=xn+an−1xn−1+⋯+a0 be a monic polynomial with coefficients in Z and a0=0
If p(x) has a zero in Q, then p(x) also has a zero α∈Z
Furthermore, α∣a0
Proof
Example (textbook)
Proving p(x)=x4−2x3+x+1∈Q[x] is irreducible using Gauss’s lemma and its corollary
Assume p(x) is reducible, then either p(x) has a linear factor (i.e. p(x)=(x−α)q(x), where degq(x)=3) or p(x) has two quadratic factors
Case 1 (linear factor)
If p(x) has a linear factor in Q[x], then it has a zero in Z that divides 1 (see corollary 17.15)
A more formal way of considering this problem is to examine fractions in terms of equivalence relations
We can think of Q as ordered pairs Z×Z, where a quotient qp is written as (p,q), where q=0
We consider two pairs, (a,b) and (c,d), to be equivalent if ad=bc
So we consider the following set S in terms of any integral domain D, and relation on S
S={(a,b):a,b∈D and b=0}
Define a relation on S by (a,b)∼(c,d) if ad=bc
Lemma 18.1
The relation ∼ between elements of S is an equivalence relation
Proof
Field of fractions FD
We’ll denote the set of equivalence classes on S by FD, now we need to define the operations of addition and multiplication on FD
Recall how fractions are added and multiplied
ba+dc=bdad+bc
ba⋅dc=bdac
It seems reasonable to define addition and multiplication on FD in a similar manner
If we denote the equivalence class of (a,b)∈S by [a,b], then we define the operations of addition and multiplication on FD by
[a,b]+[c,d]=[ad+bc,bd]
[a,b]⋅[c,d]=[ac,bd]
Lemma 18.2
The operations of addition and multiplication on FD are well-defined
Proof
Lemma 18.3
The set of equivalence classes of S, FD, under the equivalence relation ∼, together with the operations of addition and multiplication defined by [a,b]+[c,d]=[ad+bc,bd] and [a,b]⋅[c,d]=[ac,bd] is a field
Proof
The field FD is called the field of fractions or field of quotients of the integral domain D
By definition, D is a subring of FD, with r↦1r
Properties of FD
Theorem 18.4
Let D be an integral domain, then D can be embedded in a field of fractions FD, where any element FD can be expressed as the quotient of two elements in D
Furthermore, the field of fractions FD is unique in the sense that if E is any field containing D, then there exists a map ψ:FD→E giving an isomorphism with a subfield of E such that ψ(a)=a for all elements a∈D, where we identify a with its image in FD
Proof
Example (textbook)
Since Q is a field, Q[x] is an integral domain
The field of fractions of Q[x] is the set of all rational expressions p(x)/q(x), where p(x),q(x)∈Q[x] and q(x)=0
This field is denoted Q(x)
Corollary 18.6
Let F be a field of characteristic zero, then F contains a subfield isomorphic to Q
Corollary 18.7
Let F be a field of characteristic p, then F contains a subfield isomorphic to Zp
Example (lecture)
(x3+x+1) is a maximal ideal in F2[x]
So F2[x]/(x3+x+1) is a field
F2[x]/(x3+x+1) is actually F8 (i.e. a field of size 8) because it has eight elements
Note
F8=Z8
F8 just means a field of size 8
char F8=2 because it’s in F2[x]
18.2 - Factorization in integral domain
May 5 Textbook Notes
Definitions
Given an integral domain
a divides b (a∣b) - ∃c∈R,b=ac
Unit - an element with a multiplicative inverse
Associates - two elements a,b∈R if there exists a unit u∈R such that a=ub
Irreducible (element) - when a nonzero element p∈D that’s not a unit provided that whenever p=ab, either a or b is a unit
Prime (element) - a nonzero element such that whenever p∣ab, either p∣a or p∣b
i.e. p is prime if (p) is a prime ideal
Example
Prime and irreducible elements don’t always coincide
Let R be the subring (with identity) of Q[x,y] generated by x2, y2 and xy
Each of these elements is irreducible in R
But xy isn’t prime
Since xy∣x2y2 but xy∤x2 and xy∤y2
Unique factorization domain (UFD)
An integral domain D is a unique factorization domain (UFD) if D satisfies the following criteria
Let a∈D such that a=0 and a isn’t a unit, then a can be written as the product of irreducible elements in D
Let a=p1⋯pr=q1⋯qs, where the pi’s and the qi’s are irreducible, then r=s and there exists a π∈Sr such that pi and qπ(j) are associates for j=1,⋯,r
Examples of UFDs (or not)
Z is a UFD by the Fundamental Theorem of Arithmetic
Z[3i] isn’t a UFD
Let z=a+b3i
Let’s define ν:Z[3i]→N∪{0} such that ν(z)=∣z∣2=a2+3b2
From learning complex numbers, we know ν(z)≥0 (with equality when ν=0) and ν(zw)=ν(z)ν(w)
We know that the only units of Z[3i] are ±1
4=2⋅2=(1+3i)(1−3i)
2 is not associates with either 1+3i nor 1−3i
So Z[3i] isn’t a UFD
Similarly, Z[5i] isn’t a UFD
Principal ideal domains (PID)
Recall a principal domain generated by a∈R, where R is a commutative ring with identity, is (a)={ra:r∈R}
Principal ideal domain (PID) - an integral domain where every ideal is principal
Lemma 18.11
Let D be an integral domain and let a,b∈D, then
a∣b⟺(b)⊂(a)
a and b are associates ⟺(b)=(a)
a is a unit in D⟺(a)=D
Proof
Theorem 18.12
Let D be a PID and (p) be a nonzero ideal in D, then (p) is a maximal ideal ⟺p is irreducible
Proof
Corollary 18.13
Let D be a PID, if p is irreducible, then p is prime
Proof
Proposition
Let D be an integral domain, p is prime ⟹p is irreducible
Proof
Suppose p is irreducible and p=0, since p is irreducible, then (p)=D, which means that p isn’t a unit
Let ab∈(p), we’ll show a∈(p) or b∈(p)
ab∈(p)⟹ab=pc for some c∈D⟹p∣ab since D is an integral domain ⟹p must be an irreducible factor for either a or b
Without loss of generality, let p∣a
Then a∈(p)
Therefore (p) is a prime ideal ⟹p is prime
Proposition
Let D be UFD/PID, p is irreducible ⟺p is prime
Lemma 18.14 (ascending chain condition (ACC))
Let D be a PID, and I1,I2,⋯ be a set of ideals such that I1⊂I2⊂⋯, then there exists an integer N such that In=IN for all n≥N
Proof
Any commutative rings satisfying the ACC are called Noetherian rings
Theorem 18.15
Every PID is a UFD
Proof
Corollary 18.16
Let F be a field, then F[x] is a UFD
Proposition
Let R be a UFD, then R[x] is a UFD
Example
Z[x] is a UFD (which will be proved in the next section) but it’s not a PID
Let I={5f(x)+xg(x):f(x),g(x)∈Z[x]}=(5,x) be an ideal in Z[x]
Suppose, for contradiction, that I=(p(x)) for p(x)∈Z[x]
Since 5∈I, 5=f(x)p(x)⟹p(x) must be a constant
Since x∈I, x=p⋅g(x)⟹p=±1⟹(p(x))=Z[x]
This means that an element in Z[x], e.g. 3, is also in I
Therefore we can write 3=5f(x)+xg(x), for f,g∈Z[x]
⟹3=5f(x)+x⋅0=5f(x)⟹f(x)∈/Z[x]
Contradiction
All fields F are PIDs
F[x] for field F are PIDs
May 8 Lecture + Textbook Notes
Euclidean domains
Euclidean domain - an integral domain D such that there’s a function ν:D∖{0}→N satisfying the following conditions:
If a and b are nonzero elements in D, then ν(a)≤ν(ab)
Let a,b∈D and suppose that b=0, then there exist elements q,r∈D such that a=bq+r, and either r=0 or ν(r)<ν(b)
Euclidean valuation - the function ν:D∖{0}→N for Euclidean domain D
Examples of Euclidean domains/valuation
Absolute value of Z is a Euclidean valuation
Let F be a field, then the degree of a polynomial in F[x] is a Euclidean valuation
Theorem 18.21
Every Euclidean domain is a PID
Proof
Corollary 18.22
Every Euclidean doamin is a UFD
Factorization in D[x]
Let D be a UFD and suppose that p(x)=anxn+⋯+a1x+a0∈D[x]
The content of p(x) is a greatest common divisor of a0,⋯an
p(x) is primitive if gcd(a0,⋯,an)=1
Example
In Z[x], the polynomial p(x)=5x4−3x3+x−4 is a primitive polynomial since the greatest common divisor of the coefficients is 1; but q(x)=4x2−6x+8 isn’t
Theorem 18.24 (Gauss’s lemma)
Let D be a UFD and let f(x) and g(x) be primitive polynomials in D[x], then f(x)g(x) is primitive
Proof
Lemma 18.25
Let D be a UFD, and let p(x),q(x)∈D[x], then the content of p(x) is equal to the product of the contents of p(x) and q(x)
Proof
Lemma 18.26
Let D be a UFD and F its field of fractions
Suppose that p(x)∈D[x] and p(x)=f(x)g(x), where f(x),g(x)∈F[x], then p(x)=f1(x)g1(x), where f1,g1∈D[x]
Furthermore, degf(x)=degf1(x) and degg(x)=degg1(x)
Proof
Corollary 18.27
Let D be a UFD and F its field of fractions, a primitive polynomial p(x)∈D[x] is irreducible in F[x]⟺it’s irreducible in D[x]
Corollary 18.28
Let D be a UFD and F its field of fractions
If p(x) is a monic polynomial in D[x] with p(x)=f(x)g(x)∈F[x], then p(x)=f1(x)g1(x), where f1,g1∈D[x]
Furthermore, degf(x)=degf1(x) and degg(x)=degg1(x)
Theorem 18.29
If D is a UFD, then D[x] is a UFD
Proof
Corollary 18.30
Let F be a field, then F[x] is a UFD
Corollary 18.31
The ring of polynomials over the integers Z[x] is a UFD
Corollary 18.32
Let D be a UFD, then D[x1,⋯,xn] is a UFD
Remark
Every Euclidean domain is a PID and every PID is a UFD
The converse fails
Chapter 21 - Fields
21.1 - Extension fields
Extension fields
May 10 Lecture + Textbook Notes
Extension fields
Extension field - a field E is an extension field of a field F if F is a subfield of E
The field F is the base field, F⊂E
Examples (textbook)
Let F=Q(2)={a+b2:a,b∈Q}
Let E=Q(2+3) be the smallest field containing both Q and 2+3
Both E and F are extension fields of the rational numbers
E is an extension field of F
We need to show 2∈E
SInce 2+3∈E, (2+3)−1=2+31=3−2∈E
Taking linear combination of 2+3 and 3−2, we find that both 2 and 3 are in E
Let p(x)=x2+x+1∈Z2[x]
Since neither 0 nor 1 is a root of p(x), it’s irreducible over Z2
We’ll construct a field extension of Z2 containing an element α such that p(α)=0
By theorem 17.22, the ideal (p(x)) generated by p(x) is maximal
Hence Z2[x]/(p(x)) is a field
Let f(x)+(p(x)) be an arbitrary element of Z2[x]/(p(x))
By the division algorithm, f(x)=(x2+x+1)q(x)+r(x)
Therefore, f(x)+(x2+x+1)=r(x)+(x2+x+1)
The only possibilities for r(x) are 0,1,x,1+x
Consequently, E=Z2[x]/(p(x)) is a field with four elements and must be a field extension of Z2, containing a zero α of p(x)
The field Z2(α) consists of elements
Notice that α2+α+1=0
Hence, (1+α)2=1+2α+α2=α
Other calculations are accomplished in a similar manner
We summarize these computations in the following tables, which tell us how to add and multiply elements in E
Examples (lecture)
Let Q be the base field
Q(2) is the extension field of Q
R is the extension field of Q
C is the extension field of R
F(x) is the extension field of field F
From chapter 18.1, properties of FD section, we know F8=F2[x]/(x3+x+1)
F8={0,1,x,x+1,x2,x2+1,x2+x,x2+x+1}
The inverse of each element is:
0 cannot have inverse
1−1=1
x−1=x2+1
(x+1)−1=x2+x
(x2+x+1)−1=
F8 is an extension field of F2
Theorem 21.5
Let F be a field and let p(x) be a nonconstant polynomial in F[x], then there exists an extension field E of F and an element α∈E such that p(α)=0
Proof
Example (textbook)
Let p(x)=x5+x4+1∈Z2[x], then p(x) has irreducible factors x2+x+1 and x3+x+1
For a field extension E of Z2 such that p(x) has a root in E, we can let E be either Z2[x]/(x2+x+1) or Z2[x]/(x3+x+1)
Construct extension fields (lecture)
Take field F and its field of polynomials F[x]
Let f(x)∈F[x] be an irreducible polynomial
Then F[x]/(f(x)) is an extension of F
Examples
Take field Q[x]/(x2+x+1)={ax+b∣a,b∈Q}, prove it’s an extension of Q[x]
Need to show x2+x+1 is irreducible in Q[x]
If x2+x+1 has a root in Q, it has a root in Z and is a factor of 1
Factor =±1
Plugging it in, none works
x2+x+1 is irreducible in Q[x]
Turns out, Q[x]/(x2+x+1)≅Q(e2πi/3)
Using unit circle (?)
Q[x]/(x2−2)={ax+b∣a,b∈Q} is a field extension of Q[x]
Because x2−2 is irreducible in Q[x] (using Eisenstein’s criterion)
Since Q[x]/(x2+x+1) and Q[x]/(x2−2) look very similar, are they isomorphic?
How can we tell if they are isomorphic?
Do they have the same properties
Are they both integral domains
Are they both fields
For a certain question, are there solutions in both fields?
In this case, they are not isomorphic
Is there a solution for y2−2=0 in both fields?
In Q[x]/(x2−2)
We know x∈Q[x]/(x2−2)
But x2=x2(mod(x2−2))=2
So it does have a solution, y=±x
In Q[x]/(x2+x+1)
Let y=a+bx⟹(a+bx)2−2=0
a2+2abx−b2x−b2−2=0
⟹(a2−b2−2)+(2ab−b2)x=0
⟹a2−b2−2=0 and 2ab−b2=0
There are no solutions
Hence, they aren’t isomorphic
However, R[x]/(x2+1)≅R[x]/(x2+2)
Using the isomorphism theorem
Algebraic/transcendental
May 12 Lecture + Textbook Notes
Algebraic/transcendental elements
Algebraic (element over F) - an element α∈E, for extension field E over field F, where f(α)=0 for some nonzero polynomial f(x)∈F[x]
Transcendental (element over F) - an element in E, for extension field E over field F, that’s not algebraic
Algebraic extension (of field F) - an extension field E over field F where every element in E is algebraic over F
Transcendental extension (of field F) - an extension field E over field F where there exists at least an element that’s transcendental over F
If E is an extension of field F and α1,⋯,αn∈E, we denote the smallest field containing F and α1,⋯,αn as F(α1,⋯,αn)
Simple extension (of field F) - an extension field E over field F where E=F(α) for some α∈E
Examples of algebraic and transcendental elements (textbook)
Let F=Q
Both 2 and i are algebraic over Q since they are the zeros of x2−2 and x2+1 respectively
π and e are algebraic over R since they are zeros of x−π and x−e
But they are trascendental over Q
Numbers in R that are algebraic over Q are quite rare
Almost all real numbers are transcendental over Q
Let say, we pick a real number at random from the interval [0,1]
The probability that number is transcdental over Q is 100% (but not impossible)
Additionally, in many cases we do not know whether or not a particular number is transcendental
For example, it is still not known whether π+e is transcendental or algebraic
Examples of algebraic and transcendental elements (lecture)
For extension F(x) over field F
x is transcendental over F
Algebraic/transcendental numbers
Algebraic number - a complex number that’s algebraic over Q
Transcendental number - a complex number that’s transcendental over Q
Example (textbook)
We’ll show 2+3 is algebraic over Q
If α=2+3, then α2=2+3
⟹α2−2=3
⟹(α2−2)2=3
Since α4−4α2+1=0∈Q[x], it must be true that α is a zero of the polynomial
Hence, α is algebraic over Q
Theorem 21.9
Let E be an extension field of field F and α∈E, then α is transcendental over F⟺F(α) is isomorphic to F(x), the field of fractions of F[x]
Proof
Degree of field extension (lecture)
Degree of field extension E over field F - the dimension of E as a F vector space (denoted by [E:F])
Examples
Field extension Q(2) over field Q
The basis is {1,2}
⟹[Q(2):Q]=2
Extension F8={0,1,x,x+1,x2,x2+1,x2+x,x2+x+1} over field F2
Basis is {1,x,x2}
[F8:F2]=3
Extension Q(ζ3), where ζ3=e2πi/3, over field Q
We know ζ32+ζ3+1=0
⟹ζ32=−ζ3−1, so it can be written as a linear combination of ζ3 and 1
Basis is {1,ζ3}
[Q(ζ3):Q]=2
Extension F3(x) over field F3
Basis is unknown (there’s no way to know the basis)
[F3(x):F3]= countable infinity
Extension C over field R
Basis is {1,i}
[C:R]=2
Extension Q(32) over field Q
Basis is {1,32,(32)2}
[Q(32):Q]=3
Extension R over field Q
Basis is unknown
[R:Q]=∞
Minimal polynomial
May 15 Lecture Notes
Galois theory - preview (lecture)
Let’s look at field Q(2,3), and its automorphism
It’s an automorphism when the same elements of the field solve the same problem
2↦±2
3↦±3
So there are 4 automorphisms
These 4 automorphisms form a group that’s isomorphic to Z2×Z2
It’s not isomorphic to Z4 because Z4 is cyclic
Let ϕ1(2)=2 and ϕ1(3)=3
ϕ1 is the equivalent of (0,0)=e∈Z2×Z2
The order of ϕ1 in the group of automorphism is 1
Let ϕ2(2)=−2 and ϕ2(3)=3
ϕ2 is the equivalent of (1,0)∈Z2×Z2
Because ϕ22(2)=2
⟹order of ϕ2 is 2
Let ϕ3(2)=2 and ϕ3(3)=−3
ϕ3 is the equivalent of (0,1)
Let ϕ4(2)=−2 and ϕ4(3)=−3
ϕ4 is equivalent to (1,1)
Z2×Z2 has 5 subgroups
e
⟨(1,0)⟩
⟨(0,1)⟩
⟨(1,1)⟩
Z2×Z2
So does Q(2,3)
Q(2,3)
Q(2)
Q(3)
Q(6)
Q
Minimal polynomial - introduction (lecture)
Let E be extension of F
Let α∈E where α is algebraic over F
This mean ∃f(x)∈F[x] such that f(α)=0, where degf(x)≥1
f(x) isn’t unique because can just multiple f(x) by any other polynomial
Criteria of finding “good” f(x)
Degree of f(x) is as small as possible
f(x) can be multiplied by a unit in F[x] (i.e. a nonzero constant)
For simplicity, we can assume f(x) is monic / adjust f(x) to be monic
May 19 Lecture + Textbook Notes
Minimal polynomial
Theorem 21.10 (minimal polynomial)
Let E be an extension field of a field F and α∈E with α algebraic over F, then there exists a unique irreducible monic polynomial (called the minimal polynomial for α over F) p(x)∈F[x] of smallest degree such that p(α)=0
If f(x) is another polynomial in F[x] such that f(α)=0, then p(x)∣f(x)
Proof
Degree of (algebraic element) α over F - the degree of the minimal polynomial p(x) for α over F
Example
Let f(x)=x2−2 and g(x)=x4−4x2+1
These polynomials are the minimal polynomials for 2 and 2+3 respectively
Examples of calculating minimal polynomial (lecture)
α=2+3
α2=2+3
(α2−2)=3
α4−4α2+1
f(x)=x4−4x2+1
The minimal polynomial has to be degree 4 because [Q(α):Q]=4
f(x) has no root (by the rational root theorem) and has no degree 2 factors, so f(x) is irreducible ⟹minimal polynomial for α is f(x)
α=2+3
The basis for Q(α)=Q(2+3) is {1,2,3,6}
So we already know (without needing to calculate) that the degree of the minimal polynomial will be 4
α2=2+26+3
α2−5=26
(α2−5)2=24
α4−10α2+1=0
f(x)=x4−10x+1 is the minimal polynomial
Proposition
The degree of the minimal polynomial for α over F equals to [F(α):F] for field F
Proposition
The degree of the minimal polynomial equals to the number of solutions
Example (lecture)
32, 32⋅ζ3, and 32⋅ζ32 are solutions to x3−2
Proposition 21.12
Let E be the field extension of F and α∈E be algebraic over F, then F(α)≅F[x]/(p(x)), where p(x) is the minimal polynomial of α over F
Proof
Theorem 21.13
Let E=F(α) be a simple extension of F, where α∈E is algebraic over F
Suppose the degree of α over F is n, then every element β∈E can be expressed uniquely in the form β=b0+b1α+⋯+bn−1αn−1, for bi∈F
Proof
Example
Since x2+1 is irreducible over R, (x2+1) is a maximal ideal in R[x]
So E=R[x]/(x2+1) is a field extension of R that contains a root of x2+1
Let α=x+(x2+1), we can identify E with the complex numbers
By proposition 21.12, E is isomorphic to R(α)={a+bα:a,b∈R}
We know that α2=−1 in E since α2+1=x2+1=0
Hence, we have an isomorphism of R(α) with C defined by the map that takes a+bα to a+bi
Finite extensions
May 19 Lecture Notes
Finite extensions (lecture)
Theorem
Let L be an extension of field K, and E be an extension of L, if L,K are all finite extensions, then [E:K]=[E:L][L:K]
Theorem
Let E be a finite extension of F, then E is an algebraic extension
Proof
Suppose [E:F]=n, and α∈E
Let’s look at 1,α,α2,⋯,αn, there are n+1 elements
∃c0,c1,⋯,cn∈F such that c0(1)+c1(α)+⋯+cn(αn)=0
⟹c0+c1x+⋯+cnxn∈F[x]
Thus E is algebraic over F
Corollary
Algebraic elements are closed under addition and multiplication
Proof
If α,β are algebraic over field F, then F(α),F(β) are finite extensions over F
⟹F(α,β) is a finite extension over F⟹F(α,β) is an algebraic extension over F
Hence α+β,αβ∈F(α,β) are algebraic
May 22 Lecture + Textbook Notes
Finite extensions
Let E be a field extension of field F
If we regard E as a vector space over F, then we can use the knowledge of linear algebra to study fields
We can think of elements in field E as vectors and elements in field F as scalars
Addition in E is adding vectors, and multiplying an element in E with element in F is scalar multiplication
We can think of E is a finite dimensional vector space over F
If an extension field E of field F is a finite dimensional vector space over F of dimension n, then we say that E is a finite extension of degree n over F (denoted [E:F]=n)
Example
Let’s determine an extension field of Q containing 3+5
It’s easy to determine it has minimal polynomial of x4−16x2+4
From this, we can conclude that [Q(3+5):Q]=4
The basis for Q(3+5) over Q is {1,3,5,15}
Let’s compute a basis for Q(35,5i)
We know 5i∈/Q(35), so a basis for Q(35,5i) over Q(35) is {1,5i}
[Q(35,5i):Q(35)]=2
We also know that {1,35,325} is a basis for Q(35) over Q
Hence a basis for Q(35,5i) over Q is
{1,5i,35,65i,325,(65)5i}
Theorem 21.15
Every finite extension field E of field F is an algebraic extension
Proof
Remark
The converse of theorem 21.15 is false
i.e. E over F is algebraic doesn’t imply E is finite
Proof (lecture)
We have ∞-dimension extensions that are algebraic
Theorem 21.17
If E is a finite extension of F and K is finite extension of E, then K is a finite extension of F and [K:F]=[K:E][E:F]
Proof
Corollary 21.18
If Fi is a field, for i=1,⋯,k, and Fi+1 is a finite extension of Fi, then Fk is a finite extension of F1 and [Fk:F1]=[Fk:Fk−1]⋯[F2:F1]
Corollary 21.19
Let E be an extension field of F, if α∈E is algebraic over F with minimal polynomial p(x) and β∈F(α) with minimal polynomial q(x), then degq(x)∣degp(x)
Proof
Theorem 21.22
Let E be a field extension of F, then the following statements are equivalent
E is a finite extension of F
There exists a finite number of algebraic elements α1,⋯,αn∈E such that E=F(α1,⋯,αn)
There exists a sequence of fields E=F(α1,⋯,αn)⊃F(α1,⋯,αn−1)⊃⋯⊃F(α1)⊃F, where each field F(α1,⋯,αi) is algebraic over F(α1,⋯,αi−1)
Algebraic closure
May 22 + 24 Lecture + Textbook Notes
Algebraic closure
Theorem 21.23
Let E be an extension field of F, the set of elements in E that are algebraic over F form a field
Proof
Corollary 21.24
The set of all algebraic numbers forms a field, that is, the set of all complex numbers that are algebraic over Q makes up a field
Algebraic closure (of a field F in expansion field E) - the field consisting of all elements in E that are algebraic over F
In other words, for F(α1,α2,⋯), we look at all the irreducible polynomials and adjoin all the roots to out field extension
The outcome is called algebraic closure of F (denoted F)
Example (lecture)
Q∋2,3,2+3,32,i,32+i,ζ3,⋯
Q∋π,e,x
Q over Q is algebraic
Q is not finite
Is Fp finite or infinite?
Hint
If you can find polynomials with no repeated roots of arbitrary degree, it shows that Fp is infinite
Let f(x)=xpn−x
From earlier homework:
If gcd(f(x),f′(x))=1⟹f(x) has no repeated roots
f′(x)=pnxpn−1−1=−1 (since charFp=p)
⟹gcd(f(x),f′(x))=1
Because Fp contains all roots of f(x), which are distinct, n is not fixed
⟹∣Fp∣=∞
Algebraically closed - when, for a field F, every nonconstant polynomial in F[x] has a root in F
Theorem 21.25
A field F is algebraically closed ⟺ every nonconstant polynomial in F[x] factors into linear factors over F[x] (i.e. F=F)
Proof
Corollary 21.26
An algebraically closed field F has no proper algebraic extension E
Proof
Corollary
If field F is algebraically closed, then F is infinite
Proof
Let F be a field, then char p=0 or p
If char F=0, then F⊃Q⟹F is infinite
If char F=p, then F⊃Fp
Since F=F, F⊃Fp
From a previous example, we showed that Fp is infinite
Therefore F is infinite
Theorem 21.27
Every field F has a unique algebraic closure
i.e. F is unique up to isomorphism
i.e. if E1,E2 are algebraic closures of F, then E1≅E2
Theorem 21.28 (fundamental theorem of algebra)
The field of complex numbers is algebraically closed
21.2 - Splitting fields
May 24 + 26 Lecture + Textbook Notes
Splitting fields - introduction
For field F, let p(x)∈F[x] be a nonconstant polynomial, we would like to know
Does there exist an extension E over F such that all the roots of p(x) are in E
i.e. does there exists an extension E such that p(x) factors into a product of linear polynomials?
If such extension exists, what is the smallest extension?
Spltting field - existence
Let F be a field and p(x)=a0+a1x+⋯anxn∈F[x] be a nonconstant polynomial
Splitting field (of p(x)) - an extension field E over F such that there exists elements α1,⋯,αn∈E such that E=F(α1,⋯,αn) and p(x)=(x−α1)⋯(x−αn)
A polynomial p(x)∈F[x] splits in E if it’s the product of linear factors in E[x]
Examples (textbook)
Let p(x)=x4+2x2−8∈Q[x]
p(x) has irreducible factors x2−2 and x2+4
Therefore the field Q(2,i) is a splitting field for p(x)
Let p(x)=x3−3∈Q[x]
Then p(x) has a root in Q(33)
However, this field is not a splitting field for p(x) since the complex cube roots of 3 are not in Q(33), where x=2−33±(63)5i
From lecture - instead, Q(33,ζ3) is the splitting field
Example (lecture)
Let f(x)=x3+x+1∈F2[x]
f(x) has a root in F8, which is x⟹x3+x+1=0
x denotes the element x∈F8, this is to avoid confusion between the element x∈F8 and the input x of f(x)
In fact, F8 is the splitting field of f(x)
If all roots of f(x) are in F8, then F2(all roots)⊆F8
But F8=F2(one root)⊂F2(all roots)
Since char F8=2
We can calculate (x3+x+1)2=x6+x2+1=(x2)3+x2+1=0
Therefore x2 is a root of f(x)
Similarly, x4=x2+x is also a root of f(x)
Since F8 contains all three roots of f(x), F8 is the splitting field of f(x) over F2
Theorem 21.31
Let p(x)∈F[x] be a nonconstant polynomial, then there exists a splitting field E for p(x)
Proof
Splitting field - uniqueness
Now the question of the uniqueness of a splitting field arises
Yes it is unique such that, given two splitting field K and L of polynomial p(x)∈F[x], there exists an isomorphism ϕ:K→L that perserves F
Lemma 21.32
Let ϕ:E→F be an isomorphism of fields, K be an extension field of E and α∈K be algebraic over E with minimal polynomial p(x)
Suppose L is an extension field of F such that β is a root of the polynomial in F[x] obtained from p(x) under the image of ϕ
Then ϕ extends to a unique isomorphism ϕ:E(α)→F(β) such that ϕ(α)=β and ϕ agrees with ϕ on E
Proof
Theorem 21.34
Let ϕ:E→F be an isomorphism of fields, p(x) be a nonconstant polynomial in E[x], and q(x) is the corresponding polynomial in F[x] under the isomorphism
If K is a splitting field of p(x) and L is a splitting field of q(x), then ϕ extends to an isomorphism ψ:K→L
Proof
Corollary 21.36
Let p(x) be a polynomial in F[x], then there exists a splitting field K of p(x) that is unique up to isomorphism
Example (lecture)
Consider p(x)=x3−2
Q(32)≅Q(32,ζ3)
Chapter 22 - Finite fields
22.1 - Structure of a finite field
May 21 Lecture + Textbook Notes
Characteristic of finite field
Recall that a field F has either has characteristic p, where p is prime, or 0 from theorem 16.19
Suppose that F is a finite field with n elements, then nα=0 for all α∈F
Consequently, the characteristic of F must be p, where p is a prime dividing n
Proposition 22.1
If F is a finite field, then the characteristic of F is p, where p is prime
Proposition 22.2
If F is a finite field of characteristic p, then the order of F is pn for some n∈N
Proof
Lemma 22.3 - Freshman’s dream
Let p be prime and D be an integral domain of characteristic p, then apn+bpn=(a+b)pn for all positive integers n
Proof
Seperable extensions + Galois field
Let F be a field, a polynomial f(x)∈F[x] of degree n is separable if it has n distinct roots in the splitting field of f(x)
i.e. f(x) is separable when it factors into distinct linear factors over the splitting field of f
An extension E of F is a separable extension of F if every element in E is the root of a separable polynomial in F[x]
Example
Let f(x)=x2−2∈Q[x]
It’s separable over Q since it factors as (x−2)(x+2)
In fact, Q(2) is a separable extension of Q
Let α=a+b2 be any element in Q(2)
If b=0, then α is a root of x−a
If b=0, then α is the root of the separable polynomial x2−2ax+a2−2b2=(x−(a+b2))(x−(a−b2))
There’s an easy test to determine the separability of any polynomial
Let f(x)=a0+⋯+anxn∈F[x], and f′(x)=a1+2a1x+⋯+nanxn−1
Lemma 22.5
Let F be a field and f(x)∈F[x], then f(x) is separable ⟺f(x) and f′(x) are relatively prime
Proof
Theorem 22.6 - Galois field
For every prime p and every positive integer n, there exists a finite field with pn elements (called the Galois field of order pn, denoted GF(pn) or Fpn)
Furthermore, any field of order pn is isomorphic to the splitting field of xpn−x over Zp
i.e. Fpn=Fp(all roots of xpn−x) is the set of all roots of xpn−x
Proof
Theorem 22.7
Every subfield of the Galois field GF(pn) has pm elements, where m divides n
Conversely, if m∣n for m>0, then there exists a unique subfield of GF(pn) isomorphic to GF(pm)
i.e. Fpm⊂Fpn⟺m∣n
Proof
Example
The lattice of subfields of GF(p24) is
Theorem 22.10
If G is a finite subgroup of F∗, where F∗ is the multiplicative group of nonzero elements of F (i.e. F∗ is F∖{0}), then G is cyclic
Proof
Corollary 22.11
The multiplicative group of all nonzero elements of a finite field is cyclic
Corollary 22.12
Every finite extension E of a finite field F is a simple extension of F
Proof
Let α be the generator for the cyclic group E∗ of nonzero elements of E, then E=F(α)
Example
Proposition
Fpn∗=⟨α⟩⟹Fpn=Fp(α)
Proof
Fp(α)⊆Fpn
Since Fp⊂Fpn and α∈Fpn, Fp(α)⊂Fpn
Fpn⊆Fp(α)
If x∈Fpn, x is either 0 or a unit
i.e. x=0 or x∈Fpn∗
⟹x=αn, for some n∈Z
⟹x∈Fp(α)
Example
Let F64∗=⟨α⟩, find F4=F2(?)
Let’s see if F4=F2(α16)
We know the order of α is 63
Order of α16=gcd(16,∣α∣)∣α∣=63
⟹⟨α16⟩=F64∗
⟹F2(α16)=F64
So F4=F2(α16)
Let F2(β)=F4, we want β as a generator of F4∗ (by the proposition)
Order of β should be 3, and we have that order of α=63
⟹β=α21 or α42
⟹F2(α21)=F4
We know F8⊂F64, find γ∈F64 such that F2(γ)=F8
Order of γ should be 7, and we know that order of α=63
⟹γ=α9,α18,α27,α36,α45,α54
F2(α9)=F8
Chapter 23 - Galois theory
23.1 - Field automorphisms
Jun 2 Lecture + Textbook Notes
Field automorphism
Proposition 23.1
The set of all automorphisms of a field F is a group under composition of functions
Proof
Proposition 23.2
Let E be a field extension of F, then the set of all automorphisms of E that fix F elementwise is a group
i.e. the set of all automorphisms σ:E→E such that σ(α)=α for all α∈F is a group
Proof
Let E be a field extension of F
The full group of automorphisms of E is denoted Aut(E)
Galois group (of E over F) - the group of automorphism of E that fix F elementwise (denoted G(E/F) in the textbook or Aut(E/F))
Aut(E/F)={σ∈Aut(E):σ(α)=α for all α∈F}
If f(x)∈F[x] and E is the splitting field of f(x) over F, then we define the Galois group of f(x) to be Aut(E/F)
Examples (textbook)
Let σ:a+bi↦a−bi
σ is complex conjugation
It’s an automorphism of the complex numbers
Since σ(a)=σ(a+0i)=a−0i=a, the automorphism defined by complex conjugation must be in Aut(C/R)
Consider fields Q⊂Q(5)⊂Q(3,5)
For a,b∈Q(5), σ(a+b3)=a−b3 is an automorphism of Q(3,5), leaving Q(5) fixed
Similarly, τ(a+b5)=a−b5 is an automorphism of Q(3,5), leaving Q(3) fixed
The automorphism μ=στ moves both 3 and 5
So {id,σ,τ,μ} is the Galois group of Q(3,5) over Q
The following table shows that this group is isomorphic to Z2×Z2
We may also regard Q(3,5) as a vector space over Q that has basis {1,3,5,15}
It’s no coincidence that ∣Aut(Q(3,5)/Q)∣=[Q(3,5):Q]=4
Examples (lecture)
Following Aut(C/R), we have ϕ:C→C
ϕ(a+bi)=ϕ(a)+ϕ(b)ϕ(i)
Since a,b∈R, and they have to be fixed, ϕ(a+bi)=a+bϕ(i)
So we only need to know what ϕ does to i (not necessarily limited to conjugation)
Since i is a root of x2+1, then ϕ(i)=±i
So Aut(C/R)={id,z↦z}
Similarly, for Aut(Q(2)/Q), we only need to know what ϕ does to 2
Since 2 is a root of x2−2, ϕ(2)=±2
So Aut(Q(2)/Q)={id,a+b2↦a−b2}
Similarly, for Aut(Q(3)/Q), we only need to know what ϕ does to 3
Since 3 is a root of x2−3, ϕ(3)=±3
So Aut(Q(3)/Q)={id,a+b3↦a−b3}
So we just need to know what ϕ does to the generator
Let’s look at Aut(Q(32/Q)
ϕ(32) is a root of x3−2
Since the other two roots are complex, then ϕ(32)=32
So Aut(Q(32/Q)={id}
Proposition 23.5
Let E be a field extension of F and f(x)∈F[x], then any automorphism in Aut(E/F) defines a permutation of the roots of f(x) that lie in E
Proof
⟹
Let E be an algebraic extension of a field F
Two elements α,β∈E are conjugate over F if they have the same minimal polynomial
Example
In the field Q(2), the elements 2 and −2 are conjugate over Q
Proposition 23.6 (converse of prop. 23.5)
If α and β are conjugate over F, then there exists an isomorphism σ:F(α)→F(β) such that σ is the identity when restricted to F
Theorem 23.7
Let f(x) be a polynomial in F[x] and suppose that E is the splitting field for f(x) over F
If f(x) has no repeated roots, then ∣Aut(E/F)∣=[E:F]
Proof
Example (textbook)
Corollary 23.9
Let F be a finite field with a finite extension E such that [E:F]=k, then Aut(E/F) is cyclic of order k
Proof
Example (textbook)
Note - distinction between Aut(E/F) and G(E/F) (from lecture)
Aut(E/F) refers to the group of automorphisms of E that fixes F
Extension E of field F is a Galois extension if it’s a separable extension of F and a splitting field of some polynomial f(x)∈F[x]
If E is Galois, then ∣Aut(E/F)∣=[E:F], and Aut(E/F) would be called the Galois group that’s denoted by Gal(E/F) or G(E/F)
Examples of Galois extension (or not) (lecture)
Q(2) over Q - Galois
Q(2,3) over Q - Galois
Q(32) over Q - not Galois
Q(32,ζ3) over Q - Galois
Theorem
Let E be a finite extension of field F, E is a Galois extension of F⟺∣Aut(E/F)∣=[E:F]
Jun 5 Lecture + Textbook Notes
Separable extensions
Many results from the previous section are based on the fact that a polynomial f(x)∈F[x] has no repeated roots in its splitting field
Let E be the splitting field of a polynomial f(x)∈F[x]
Suppose that f(x) factors over E as f(x)=(x−α1)n1(x−α2)n2⋯(x−αr)nr=∏i=1r(x−αi)ni
The multiplicity of a root αi of f(x) is ni
Simple root - a root with multiplicity 1
An extension E of F is a separable extension of F if every element in E is the root of a separable polynomial in F[x]
Proposition 23.12
Let f(x) be an irreducible polynomial over F, if the characteristic of F is 0, then f(x) is separable
If the characteristic of F is p and f(x)=g(xp) for some g(x)∈F[x], then f(x) is also separable
Proof
Primitive element
We want to know, given a field extension E of F, if it’s possible to find an element α∈E such that E=F(α)
In this case, α is called the primitive element
Examples
Q(3,5)=Q(3+5)
Q(35,5i)=Q(65i)
Corollary 22.12 tells us there exists a primitive element for any finite extension of a finite field
Theorem 23.13 - primitive element theorem
Let E be a finite separable extension of field F, then there exists an α∈E such that E=F(α)
Proof
23.2 - The fundamental theorem
Jun 7 Lecture + Textbook Notes
Normal extension
Proposition 23.14 - fixed field
Let {σi:i∈I} be a collection of automorphisms of a field F, then F{σi}={a∈F:σi(a)=a for all σi} is the fixed field of {σi} and is a subfield of F
Proof
Corollary 23.15
Let F be a field and let G be a subgroup of Aut(F), then FG={α∈F:σ(α)=α for all σ∈G} is a subfield of F
Example (textbook)
Let σ:Q(3,5)→Q(3,5) be the automorphism where σ(3)=−3
Then Q(5) is the subfield of Q(3,5) left fixed by σ
Proposition 23.17
Let E be a splitting field over F of a separable polynomial, then EG(E/F)=F
Proof
Lemma 23.18
Let G be a finite group of automorphisms of E and let E=FG, then [E:F]≤∣G∣
Proof
Let E be an algebraic extension of F
If every irreducible polynomial in F[x] with a root in E has all of its roots in E, then E is called the normal extension of F
i.e. every irreducible polynomial in F[x] containing a root in E is the product of linear factors in E[x]
Theorem 23.19
Let E be a field extension of F, then the following statements are equivalent
E is a finite, normal, separable extension of F
E is a spltting field over F of a separable polynomial
F=EG for some finite group G of automorphisms of E
Proof
Corollary 23.20
Let K be a field extension of F such that F=KG for some finite group of automorphisms G of K, then G=G(K/F)
Proof
Example (textbook)
Jun 9 Lecture + Textbook Notes
Fundamental theorem of Galois theory
Theorem 23.23 - Fundamental theorem of Galois theory
Let F be a finite field or a field of characteristic zero
If E is a normal extension of F with Galois group G(E/F), then the following statements are true
The map K↦G(E/K) is a bijection of subfields K of E containing F with the subgroups of G(E/F)
If F⊂K⊂E, then
[E:K]=∣G(E/K)∣
[K:F]=[G(E/F):G(E/K)]
F⊂K⊂L⊂E⟺{id}⊂G(E/L)⊂G(E/K)⊂G(E/F)
K is a normal extension of F⟺G(E/K) is a normal subgroup of G(E/F)
